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zubka84 [21]
3 years ago
11

What is the evaluation?

Physics
1 answer:
natita [175]3 years ago
7 0
These are listed on the map, but A & B are 900', C is 1500', and D is 1000'
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What is the rotational kinetic energy of the Earth about the Sun? Assume the earth is a uniform sphere, mass of the Earth is 5.9
Thepotemich [5.8K]

Answer:

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w²  (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R²  (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

K = 2.66x10^33 J

4 0
3 years ago
A less common type of radioactive particle emitted is the positron. If potassium-38 forms argon-38 by positron emission, the num
aleksandr82 [10.1K]
Potassium goes to argon so that a proton in the K nucleus "vanishes" into a positron (positive electron ?) but keeps the same atomic mass at 38. ? I'd say that the number of the positron was 0, and the charge was +1 electronic charge.
4 0
2 years ago
Can someone help? pls
Vladimir79 [104]

Answer:

I do believe its B because its parallel

Explanation:

8 0
2 years ago
Read 2 more answers
Draw an energy chain diagram to show energy transformations for this event:
larisa86 [58]

The energy chain diagram in this case is chemical energy >> energy conversion >> motion (kinetic) energy + thermal energy.

<h3>What is an energy chain diagram?</h3>

An energy chain diagram is a graphic representation indicating the conversion between different types of energies.

Kinetic energy is a type of motion (movement) energy generated by using potential (stored) energy.

Chemical energy (in this case, the fuel of the car) is a type of energy that is stored to perform work.

Learn more about kinetic energy here:

brainly.com/question/25959744

8 0
2 years ago
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
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